Đáp án:
\(a=0,8m/{{s}^{2}}\)
Giải thích các bước giải:
\({{v}_{1}}=5m/s;AB;{{t}_{1}}=5s;{{t}_{2}}=5s;BC-AB=20m\)
ta có: Quãng đường:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & AB={{v}_{1}}.{{t}_{1}}+\frac{1}{2}.a.t_{1}^{2}=5.5+\dfrac{1}{2}.a{{.5}^{2}}=25+12,5a \\
\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ } & AC={{v}_{1}}.({{t}_{1}}+{{t}_{2}})+\dfrac{1}{2}.a.{{({{t}_{1}}+{{t}_{2}})}^{2}} \\
\text{ }\!\!~\!\!\text{ } & =5.10+\dfrac{1}{2}.a{{.10}^{2}}=50+50a \\
\end{array}\)
Mà:
\(\begin{align}
& AC=AB+BC \\
& \Rightarrow BC=AC-AB=50+50a-25-12,5a=25+37,5a \\
\end{align}\)
Theo đầu bài:
\(\begin{align}
& BC-AB=20m\Leftrightarrow 25+37,5a-25-12,5a=20 \\
& \Rightarrow a=0,8m/{{s}^{2}} \\
\end{align}\)
